Show that the function `f: R_0->R_0` , defined as `f(x)=1/x` , is one-one onto, where `R_0` is the set of all non-zero real numbers. Is the result true, if the domain `R_0` is replaced by `N` with co-domain being same as `R_0` ?

Explanation

It is given that `f:R->R` is
defined by `f(x)=1/x`

For,one-one,

`f(x)=f(y)`

`=>1/x​=1/y​`

`=>x=y`

`therefore`f is one-one.

let's check for onto:

It is clear that for y∈R⋅,
there exists `x=1/y` such that
`f(x)=1/(1/y)​=y`

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